From: Alfred Nykolyn <apn@istar.ca>
Subject: Re: A Geometry problem
Date: Thu, 22 Nov 2001 13:47:03 -0700
Newsgroups: sci.math
Summary: Adventitious angles problems (solving triangles with "unexpectedly" integral solutions)

Here is a little twist to the Adventious Angles problem which appeared
in Kvant some time ago:

[Correction]
Given an isosceles triangle ABC in which AB is equal to AC and the angle
BAC equals 20 degrees.  (so BC is the base of the triangle)
A point D is defined on AC so that AD is equal to BC. What is the size
of angle DBC?
==============================================================================

From: Lewis Mammel <l.mammel@worldnet.att.net>
Subject: Re: A Geometry problem
Date: Fri, 23 Nov 2001 07:02:21 GMT
Newsgroups: k12.ed.math,sci.math,uk.education.maths

Dave Wilson wrote:
> 
> In article <dkflvtgpl0m3qai4rirlpl9qgro491sl22@4ax.com>,
> "Zarang" <Zarang@btinternet.com> wrote:
> 
> >Many thanks to all who replied to my posting. I have now  completed the
> >problem.
> >I had not realised  that it had quite a history and that under the title of
> >"Adventitious angles " there were considerable web references to the
> >problem. I am much obliged to all who either replied to the group or
> >e-mailed me the solution.
> 
> Do, please, share some of the details of approaches, and, perhaps, some of
> the useful URLs (I know, now, that I can chuck AA into a search engine, but
> your thoughts on useful ones will assist)


The first ( top-most ) google match on [ adventitious angles ] :
http://www.dcs.st-and.ac.uk/~ad/mathrecs/advent/advent.html
gives a link to a geometric proof, and mentions that "a number of
trig. solutions are possible."

I'll mention the one that occurred to me, as it's fairly simple:

By applying the law of sines to the triangle APQ we get

AP/AQ = sin x / sin y ; x = angle AQP, y = angle APQ

by applying the law of sines to triangle APC we get

AP/AC = sin 30 / sin 130

and correspondingly with triangle AQC we get

AQ/AB = sin 20 / sin 140

then angle BQP = 30 iff x = 110, y = 50 iff

sin 110 / sin 50 = sin 30 * sin 140 / sin 20 / sin 130 iff
sin 70 / sin 50 = sin 30 * sin 40 / sin 20 / sin 50 iff
cos 20 * sin 20 / sin 30 = sin 40

but since sin 30 = 1/2, this is true by the double angle formula.

Lew Mammel, Jr.

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==============================================================================

From: "Zarang" <Zarang@btinternet.com>
Subject: Re: A Geometry problem
Date: Fri, 23 Nov 2001 16:54:17 GMT
Newsgroups: k12.ed.math,sci.math,uk.education.maths

This posting seems to have aroused considerable interest and I am grateful
to all of those who either responded  to the groups or to me by e-mail.
I should especially like to thank Dave Rusin and R.S Tiberio for their
useful references or advice.

It seems that this problem was originally proposed by  F.M. Langley and
belongs to a set of problems known as "Adventitious Angles".

A Google search  found 1,110 references to Adventitious Angles.  However the
most useful sites  which contain solutions or valuable clues are:-

http://cmgm.stanford.edu/~ahmad/solaug96.html#solution4

http://cmgm.stanford.edu/~ahmad/puzaug96.html

http://www.dcs.st-and.ac.uk/~ad/mathrecs/advent/soln1.html

I hope this information will assist those who are searching for solutions.

Zarang.


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